30-Kinetics-Integrated-HLife-Ans

30-Kinetics-Integrated-HLife-Ans - CHEMICAL KINETICS...

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CHEMICAL KINETICS INTEGRATED RATE LAWS AND HALF-LIFE GCHEM II LAB CONCEPT PROBLEMS INTEGRATED RATE LAW: 1) Decomposition of dimethyl ether (CH 3 ) 2 O at 410 o C is first order with a rate constant of 6.8 x 10 -4 sec -1 . If the initial concentration of the dimethyl ether was 0.457 M, what is the concentration after 125 seconds ? 1 st order then ln([A F ]/[A I ) = – kt ln([A F ]/[.457] = (– 6.8 x 10 –4 sec -1 )*(125) ln([A F ]/[.457] = – 0.085 [A F ]/[.457] = e - 0. 085 [A F ] = (0.9185)*(.457) [A F ] = 0.0419 [A] = 0.42M 2) The equation and rate law for hydrolysis of sucrose is as follows: C 12 H 22 O 11 C 6 H 12 O 6 Rx Rate = k [C 12 H 22 O 11 ] 2 . After 2.57 hours at 29.5 o C, the sucrose concentration decreased from 0.0146 M to 0.0132 M. Calculate the value of rate constant with units ? RxRate = k[C 12 H 22 O 11 ] 2 2 nd order reaction 1/[A F ] – 1/[A I ] = kt 1/.0132 – 1/.0146 = (k) * (2.57) (75.76 – 68.49) = (k)*(2.57) k = (7.27)/2.57
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30-Kinetics-Integrated-HLife-Ans - CHEMICAL KINETICS...

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